Question

A student has a resistance wire of 1 ohm. If the length of the wire is 50 cm, to what length he should stretch it uniformly so as to obtain a wire of 4 ohm resistance? Justify your answer.

Answer 1

The length that he should stretch it uniformly so as to obtain a wire of 4 ohm resistance is L = 200 cm

Here a student has a resistance wire  =  1 ohm

Length of the wire = 50 cm

Wire of resistance = 4 ohm

$R\quad =\quad \rho \times \frac { l }{ A }$

Plug the value;

$1\quad =\quad \rho \times \frac { 50 }{ A }$

Or $\frac { \rho }{ A } \quad =\quad \frac { 1 }{ 50 } ---------- (1)$

Now resistance = 4 ohm

$R\quad =\quad \rho \times \frac { L }{ A }$

Plug the values;        

$4\quad =\quad \rho \times \frac { L }{ A }$

Hence  $\frac { \rho }{ A } \quad =\quad \frac { 1 }{ 50 }$

So

$4\quad =\quad (\frac { 1 }{ 50 } )L$

Hence, L = 200 cm

Answer 2

Given that, a student has a resistance wire of 1 ohm and the length of this wire is 50 cm.

We have to find that how much he should stretch it uniformly so as to obtain a wire of 4 Ohm's resistance.

We know that,

Resistance of a wire is directly proportional to it's length and inversely proportional to the area of cross-section. Where rho (p = resistivity) is a contact.

R = p l/A

Multiply and divide by length (l),

R = p l/A × l/l

R = p l²/Al [ Area × Length = Volume ]

R = p l²/V

As per given condition we are talking about two wires. So,

R1/R2 = (l1)²/(l2)²

Substitute the known values in the above formula,

1/4 = (50)²/(l2)²

(1)²/(2)² = (50)²/(l2)²

1/2 = 50/l2

l2 = 100 cm